CHAPTER 10.3

Heat of Reaction

Chemical Energy and the Universe

Warming your hands with a heat pack involves the following chemical reaction:

4Fe(s) + 3O2(g) 2Fe2O3 (s) + 1625 kJThe pack and its contents would be called the

system.Everything else is called the surroundings.The surroundings + the system = the universe.When energy flows from the system to the

surroundings, heat is released and the reaction is exothermic.

When the flow of heat is reversed, heat flows from the surroundings into the system and the reaction is endothermic.

Exothermic & Endothermic

Recall the reaction in the heat pack:

4Fe(s) + 3O2(g) 2Fe2O3 (s) + 1625 kJThe heat is on the product side—energy is

released; it is exothermic.

If energy is located on the reactant side, the reaction is endothermic as in the following reaction of a cold pack:

27 kJ + NH4NO3(s) NH4+

(aq) + NO3-

Heat of Reaction

The heat content of a system is called the enthalpy.

Change in enthalpy (or heat) is represented by ΔHrxn

ΔHrxn =H products – H reactants

ΔHrxn for exothermic reactions are always negative.

In the heat pack example:4Fe(s) + 3O2(g) 2Fe2O3 (s) ΔHrxn = - 1625 kJ

ΔHrxn for endothermic reactions are always positive.

Thermochemical Equations

A thermochemical equation is a balanced chemical equation that includes the physical states of all reactants and products, and the energy change, usually expressed in enthalpy (ΔH).

Heat Pack Example: 4Fe(s) + 3O2(g) 2Fe2O3(s) ΔH = - 1625 kJ

Coefficients refer to the number of moles. Therefore, -1625 kJ is the ΔH when 2 mol of Fe2O3 is formed from 4Fe(s) and 3O2(g).

The Rules of Thermo Equations

ΔH is directly proportional to the quantity of a substance or mass.

Example:

4Fe(s) + 3O2(g) 2Fe2O3(s) ΔH = - 1625 kJ

8Fe(s) + 6O2(g) 4Fe2O3(s) ΔH = - 3250 kJΔH for a reaction is equal in magnitude but

opposite in sign for the reverse reaction. Example:

4Fe(s) + 3O2(g) 2Fe2O3(s) ΔH = - 1625 kJ

2Fe2O3(s) 4Fe(s) + 3O2(g) ΔH = + 1625 kJ

Variations on ΔH

Energy released in burning is called heat of combustion (ΔHcomb)

Energy required to vaporize one mole of a liquid (liqgas) is called heat of vaporization (ΔHvap)

Energy required to melt one mole of a solid substance (sol liq) is called heat of fusion (ΔHfus) Because melting and vaporizing are endothermic

processes, the ΔH value is positive. What sign would the ΔH have for freezing and for

condensation? Are these processes endo- or exothermic?

Important!

Heat can be used to change the temperature of a substance

ORChange the physical state of a substance.Heat CANNOT do both at the same time.For changes of state, you calculate heat with

dimensional analysis using the ΔH as a conversion factor.

Example 1

How much heat is required to vaporize 343 g of liquid ethanol, CH3OH, at its boiling point? (ΔHvap = 38.6 kJ/mol)

Use dimensional analysis to solve for heat (q)

414 kJ

343 g 1 mol 38.6 kJ=?

32.034 g 1 mol

Example 2

How much heat is evolved when 1255 g of water condenses to a liquid at 100 °C? (Hint: ΔHvap = +40.7 kJ/mol; therefore ΔHcond = -40.7 kJ/mol)

Using dimensional analysis

-2838 kJ

1255 g 1 mol -40.7 kJ = ?

18.012 g 1 mol

Calculating the Heat of Reaction

To calculate the ΔH of a reaction, follow these simple steps

1.Balance the chemical equation2.Use the numbers given in this equation:

ΔHrxn =H products – H reactants

3.Use your balanced equation and multiply each reactant and product by the coefficient.

4.Solve

Example 3- Tying it together

Hydrogen and fluorine gas can be combined to create hydrofluoric acid. The heat of formation for H2 = 436 kJ/mol, F2 = 158 kJ/mol, and HF = 568 kJ/mol. What is the heat of this reaction?

H2 + F2 2HF

ΔHrxn = ( 436 + 158) – (2 x 568) = ?

ΔHrxn = -542 kJ/mol